the chemical equation below shows the formation of aluminum oxide (al2o3) from aluminum (al) and oxygen…

Explanation:

Step1: Determine mole - ratio

From the balanced equation $4Al + 3O_2\rightarrow2Al_2O_3$, the mole - ratio of $O_2$ to $Al_2O_3$ is $\frac{3}{2}$.

Step2: Calculate moles of $O_2$

If $n(Al_2O_3)=3.80\ mol$, then $n(O_2)=\frac{3}{2}\times n(Al_2O_3)=\frac{3}{2}\times3.80\ mol = 5.70\ mol$.

Step3: Calculate mass of $O_2$

Given the molar mass of $O_2$, $M(O_2)=32.0\ g/mol$. Using the formula $m = n\times M$, we have $m(O_2)=n(O_2)\times M(O_2)=5.70\ mol\times32.0\ g/mol = 182.4\ g\approx182\ g$.

Answer:

182 grams